Question

r(t) is a position vector. v(t) is r'(t) or the velocity vector, and a(t) is the acceleration vector. u(t) is a unit r(t) vector. Also, F(t)=ma(t)=-GMmu(t)/r^2
h(t)= r(t) x v(t)=r^2 u(t) x u'(t)
show that (v(t)xh(t))'=-GMu'(t)
somehow I keep getting GMu(t) or (a(t)xh(t)) instead...

Answer 1

By you getting a(t)xh(t), you mean you keep getting\[a \times h = - G\,M\, \dot u\] ???

Answer 2

wait sorry, it's reversed. the question wants (v(t)xh(t))'=GMu˙but i keep getting (a(t)xh(t))=-GMu' instead

Answer 3

Well, then you are fine, cause\[(v \times h)' = v' \times h = a \times h =- G\, M\, \dot u\]That's because you showed earlier, that h is constant (second law of Kepler) and thus the derivative of it is 0.

Answer 4

so i guess the book's wrong...

Answer 5

Why would it be wrong? What does the book say? o.O

Answer 6

show that (v(t)xh(t))'=GMu(t)
this is the original question... lol

Answer 7

Well .. that is what you showed, isn't it?
(sry, I don't see your problem ... please elaborate)

Answer 8

I'm trying to seek confirmation that the book is missing a particular ' sign. :P it seems so. lol thanks.